Euler's Totient function has the following property: $$ \phi(p^\alpha) = p^\alpha - p^{\alpha - 1} $$
for prime p and $ \alpha \geqslant 1 $
However the following proof demonstrates that$ \phi(n) = \frac{n}{2}$ iff $n = 2^k$ for some $ k \geqslant 1 $
Let $ n = 2^km$ where m is odd. Then $\phi(n) = 2^{k-1}\phi(m)$ which equals $\frac{n}{2}$ only if $\phi(m) = m$ that is $m = 1$ So $n = 2^k$
I have two questions, first of all, why does $\phi(n) = 2^{k-1}\phi(m)$? How is this conclusion reached in the proof?
And secondly why does this not contradict the property? Given that 2 is a prime, shouldn't the property dictate that $\phi(2^k) = 2^k - 2^{k-1}$
If
$$n=2^k\implies \frac n2=2^{k-1}=2^k-2^{k-1}\;\ldots$$
so indeed $\;\phi(n)=\cfrac n2\;$ yet no contradiction exists.